Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to Wikipedia, if the Wronskian of two functions is always zero, then they are not necessarily linearly dependent.

But it seems that if the two functions are solutions of the same homogeneous second-order linear differential equation, then the condition $W[y_1, y_2](t) = 0$ does indeed imply that they are linearly dependent.

Online, I found that if two functions are real analytic and their Wronskian is identically zero, then they are necessarily linearly dependent. But there is no reason that the solutions to a linear differential equation should be real analytic.

How can we prove that the condition $W[y_1, y_2](t) = 0$ implies the linear dependence of $y_1(t)$ and $y_2(t)$? More generally, how can we prove that the condition $W[y_1, \ldots, y_n](t) = 0$ implies the linear dependence of $y_1(t), \ldots, y_n(t)$?

share|cite|improve this question
Have you heard about Abel-Liouville's formula? – Josué Tonelli-Cueto May 10 '12 at 22:16
If $x$ is the solution to a 1 dim., $n$th order linear ode, the state $\sigma(t) = (x(t), x^{(1)}(t),...,x^{(n-1)}(t))$ encapsulates all relevant information about $x$ at the time $t$. Suppose $t'>t$, then the ode defines a linear map $\Phi(t',t)$ such that $\sigma(t') = \Phi(t',t)\sigma(t)$. Under reasonable smoothness conditions, the ode is reversible, ie, $\sigma(t)$ can be obtained by running the ode backwards starting at $\sigma(t')$, or in other words $\sigma(t) = \Phi(t,t')\sigma(t')$. The point is that linear independence at time $t$ is equivalent to linear independence at time $ t'$. – copper.hat May 10 '12 at 22:55
This is treated in Tenenbaum & Pollard's book, on page 779 and 780. The theorem starts on the very bottom of 779, maybe you can make do with just what is available in the preview on 780? The theorem states that if the functions are solutions of a homogeneous linear differential equation on some interval $I$ and if the Wronskian is identically zero on $I$, then the set of functions is linearly dependent on $I$. – process91 May 11 '12 at 4:02
@BillCook Thanks, I've read the article but, at least to my understanding, it simply implies that $W$ must be zero or nonzero everywhere. it says nowhere that if the Wronskian is 0 everywhere then the two solutions are independent – Cauchy May 11 '12 at 15:15

Suppose that $y_1(t), \ldots, y_n(t)$ are solutions of $\frac{d^n y}{dt} + p_{n-1}(t) \frac{d^{n-1} y}{dt} + \cdots + p_1(t) \frac{dy}{dt} + p_0(t) y = 0$, and suppose that their Wronskian is zero for $t = t_0$, i.e. \begin{equation*} \left| \begin{array}{cccc} y_1(t_0) & y_2(t_0) & \cdots & y_n(t_0) \\ y_1'(t_0) & y_2'(t_0) & \cdots & y_n'(t_0) \\ \vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)}(t_0) & y_2^{(n-1)}(t_0) & \cdots & y_n^{(n-1)}(t_0) \end{array} \right| = 0. \end{equation*} Then the corresponding matrix is not invertible, and the system of equations \begin{array}{c} c_1 y_1(t_0) &+& c_2 y_2(t_0) &+& \cdots &+& y_n(t_0) &=& 0 \\ c_1 y_1'(t_0) &+& c_2 y_2'(t_0) &+& \cdots &+& y_n'(t_0) &=& 0 \\ \vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\ c_1 y_1^{(n-1)}(t_0) &+& c_2 y_2^{(n-1)}(t_0) &+& \cdots &+& y_n^{(n-1)}(t_0) &=& 0 \\ \end{array} has a nontrivial solution for $c_1, c_2, \ldots, c_n$ not all zero.

Let $y(t) = c_1 y_1(t) + \cdots + c_n y_n(t)$. Because $y(t)$ is a linear combination of solutions of the differential equation, $y(t)$ is also a solution of the differential equation. Additionally, because the weights satisfy the above system of equations, we have $y_1(t_0) = y_1'(t_0) = \cdots = y_1^{(n-1)}(t_0) = 0$.

These initial conditions and the original differential equation define an initial-value problem, of which $y(t)$ is a solution. If $p_0(t), p_1(t), \ldots, p_{n-1}(t)$ are continuous, then any initial-value problem associated with the differential equation has a unique solution. Obviously $y^*(t) = 0$ is a solution of the initial-value problem; since we know that $y(t)$ is also a solution of the same initial-value problem, it follows that $y(t) = 0$ for all $t$, not just $t = t_0$.

We now have $c_1 y_1(t) + \cdots + c_n y_n(t) = 0$ for all $t$, where $c_1, \ldots, c_n$ are not all zero. Thus the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent.

Conversely, if the functions $y_1(t), \ldots, y_n(t)$ are linearly dependent, then the system of equations \begin{array}{c} c_1 y_1(t) &+& c_2 y_2(t) &+& \cdots &+& y_n(t) &=& 0 \\ c_1 y_1'(t) &+& c_2 y_2'(t) &+& \cdots &+& y_n'(t) &=& 0 \\ \vdots &+& \vdots &+& \ddots &+& \vdots &=& 0 \\ c_1 y_1^{(n-1)}(t) &+& c_2 y_2^{(n-1)}(t) &+& \cdots &+& y_n^{(n-1)}(t) &=& 0 \\ \end{array} has a nontrivial solution for every $t$, the corresponding matrix is not invertible for any $t$, and $W[y_1, \ldots, y_n](t) = 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.